Let $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $\beta \hat{i}+(1-\beta) \hat{j}$ respectively…
Let $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $\beta \hat{i}+(1-\beta) \hat{j}$ respectively be the
position vectors of the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ with respect to the origin $\mathrm{O}$. If the distance of $\mathrm{C}$ from the bisector of the
acute angle between $\mathrm{OA}$ and $\mathrm{OB}$ is $\frac{3}{\sqrt{2}},$ then the sum of
all possible values of $\beta$ is:
4
3
2
1
Solution
Since, the angle bisector of acute angle between $O A$
and $O B$ would be $y=x$
Since, the distance of $C$ from bisector $=\frac{3}{\sqrt{2}}$
$\Rightarrow\left|\frac{\beta-(1-\beta)}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}=2 \beta=\pm 3+1$
$\beta=2$ or $\beta=-1$
Hence, the sum of all possible value of $\beta=2+(-1)=1$
